complete the table by determining the balance a for p dollars invested at rate r for t years and compounded…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, and for continuous compounding, $A = Pe^{rt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $t$ is the number of years, $n$ is the number of times compounded per year. Here, $P = 4000$, $r=0.05$, and $t = 12$.

Step2: Calculate when $n = 1$

$A=4000(1 +\frac{0.05}{1})^{1\times12}=4000(1.05)^{12}\approx7103.43$.

Step3: Calculate when $n = 2$

$A = 4000(1+\frac{0.05}{2})^{2\times12}=4000(1.025)^{24}\approx7234.90$.

Step4: Calculate when $n = 4$

$A=4000(1+\frac{0.05}{4})^{4\times12}=4000(1.0125)^{48}\approx7288.18$.

Step5: Calculate when $n = 12$

$A=4000(1+\frac{0.05}{12})^{12\times12}=4000(1+\frac{0.05}{12})^{144}\approx7308.14$.

Step6: Calculate when $n = 365$

$A=4000(1+\frac{0.05}{365})^{365\times12}=4000(1+\frac{0.05}{365})^{4380}\approx7312.16$.

Step7: Calculate continuous compounding

$A = 4000e^{0.05\times12}=4000e^{0.6}\approx7288.48$.

Answer:

When $n = 4$, $A\approx7288.18$; when $n = 12$, $A\approx7308.14$; when $n = 365$, $A\approx7312.16$.